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Math Help - Find all y in [0,2π]

  1. #1
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    Find all y in [0,2π]

    Find all y in [0,2π]

    a. 2(cosx)^2-\sqrt{3}\left(\frac{1}{secx}\right) = 0
    2cos^2 - \sqrt{3}\left(\frac{\frac{1}{1}{cosx}}\right) = 0
    2cos^2 - \sqrt{3} cos x= 0
    How do I progress from here?



    b. cos^2x + cosx = sin^2x
    cos^2x+cosx-sin^2x=0

    2cos^2x+cos=0

    cosx(2cosx+1)=0

    cosx=0

    2cosx+1=0

    cosx=\frac{-1}{2}

    y = \frac\pi2;\frac{3\pi}{2}; \frac{2\pi}{3}; \frac{4\pi}{3}
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  2. #2
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    Re: Find all y in [0,2π]

    Find all y in [0,2π]-12-apr-13.png
    Thanks from Unreal
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  3. #3
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    Re: Find all y in [0,2π]

    Cos^2x + cos x = sin ^2 x
    = 1 cos^2 x
    That gives
    2 cos ^2 x + cos x -1 = 0
    (Cos x + 1)( 2cosx 1 )= 0
    Cos x = -1 OR cos x =
    Now you can proceed further
    Thanks from Unreal
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  4. #4
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    Re: Find all y in [0,2π]

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